Problem 36 - Applied Calculus

Question

If you have 100 feet of fencing and want to enclose a rectangular area up against a long, straight wall, what is the largest area you can enclose?

Step-by-Step Solution

Answer

The largest area is 1250 square feet.

Step 1: Understanding the Problem

You need to maximize the area of a rectangle that you can enclose with 100 feet of fencing along three sides, as one side is against the wall. This means only two widths and one length will use the fencing.

Step 2: Setting Up the Equation

Let's denote the length along the wall as \(x\) and the width perpendicular to the wall as \(y\). The total amount of fencing available is 100 feet, so the equation relating these dimensions is \(x + 2y = 100\).

Step 3: Express Area as a Function

The area \(A\) of the rectangle is given by \(A = x imes y\). Use the fencing equation to express \(x\) in terms of \(y\): from \(x + 2y = 100\), we have \(x = 100 - 2y\). Substitute into the area function: \(A(y) = (100 - 2y) imes y = 100y - 2y^2\).

Step 4: Finding the Maximum Area

To find the maximum area, we need to find the vertex of the quadratic function \(A(y) = 100y - 2y^2\). The vertex form of a quadratic \(ax^2 + bx + c\) has its vertex at \(y = -\frac{b}{2a}\). Here, \(a = -2\) and \(b = 100\), so \(y = -\frac{100}{2(-2)} = 25\).

Step 5: Calculate Maximum Dimensions

With \(y = 25\), substitute back into \(x = 100 - 2y\) to find \(x\): \(x = 100 - 2(25) = 50\). Therefore, the dimensions for the largest area are 50 feet along the wall and 25 feet perpendicular to it.

Step 6: Determine Maximum Area

Calculate the area using the dimensions: \(A = x imes y = 50 imes 25 = 1250\) square feet.

Key Concepts

Calculus Application

Calculus is an invaluable mathematical tool used for solving various problems, including those involving optimization. In this context, it involves finding the best possible solution under given constraints. Here, calculus helps us find the maximum enclosed area with a fixed length of fencing.

To tackle problems like this "fencing problem", we use calculus to derive and analyze functions that describe what we want to optimize. We identify critical points of these functions by utilizing derivatives. When functions reach these critical points, they are at their maximum or minimum values, which is essential for solving optimization problems.

In this particular problem, setting up the function for area and finding its derivative allows us to identify where the maximum area occurs. Thus, calculus is the key to efficiently finding the optimal solution.

Quadratic Functions

Quadratic functions have the general form \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants. These functions form a parabolic shape known as a parabola when graphed. In optimization problems, such as the fencing problem, we often deal with quadratic functions to find maximum or minimum values.

In our case, the function for area \(A(y) = 100y - 2y^2\) is quadratic and opens downward, indicating it will have a maximum point, which is the vertex of the parabola. The vertex formula \(y = -\frac{b}{2a}\) helps pinpoint the value of \(y\) where this maximum occurs. This precise evaluation is critical for determining the dimensions with the largest possible area.

Understanding the properties of quadratic functions, especially how to find the vertex, is essential to solving real-world problems efficiently.

Fencing Problem

The fencing problem is a classic optimization challenge often used in mathematical education. It serves as a practical application of creating solutions where constraints are involved. The goal here is to optimize, or in this case, to maximize the area that a given amount of fencing can enclose against an existing boundary.

With a wall as one side of the rectangle, this problem simplifies the needed equations since only three sides need fencing. You must use all your resources wisely within the given constraint, which is the fixed total length of 100 feet. Expressing one variable in terms of another by using this constraint enables the reduction of dimensions and simplifies solving with calculus.

Solving fencing problems like this one enhances critical thinking skills and demonstrates practical applications that apply in numerous fields, from agriculture to architecture.

Maximum Area Solution

Finding the maximum area involves strategically leveraging the length of fencing while adhering to constraints. The optimization process leads us to the maximum point of our quadratic area function, and that's where the area is at its greatest.

The procedure begins with translating the physical constraints into mathematical equations, followed by expressing the goal, here the area, as a function. For a maximum, we identify the vertex of the quadratic function \(A(y) = 100y - 2y^2\). Through calculus, this point is determined when \(y = 25\), yielding the length \(x = 50\).

These dimensions give a maximum enclosed area of 1250 square feet. Mastering these steps can be directly applied to alternate problems involving maximizing or minimizing resources effectively.